After some discussions with @jingli and @schwab, we realized that the peak of the wavelength distribution `\phi(\lambda)`

was not at the same position as the energy distribution `\phi(E)`

. This was because, after the change of variables, the corresponding jacobian was missing. This was properly implemented for the angular distribution.

Consider that `|\phi(E) dE| = |\phi(\lambda) d\lambda|`

, and `\lambda [\textup{~\AA}] = \sqrt{81.82/E [\mathrm{meV}]}`

, it is possible to see that `dE = |81.82 \cdot (-2) \cdot \lambda^{-3}| d\lambda`

.

So, for each bin between `\lambda_{\mathrm{min}}`

and `\lambda_{\mathrm{max}}`

, it would be necessary to divide by the integral of the jacobian, i.e. `|81.82\cdot(\lambda^{-2}_{\mathrm{min}}-\lambda^{-2}_{\mathrm{max}})|`

.

For doing this, a new `WavelengthAxis`

class was implemented to plot the distributions with the proper normalization (same idea as `AngleAxis`

). Also, the `tracks-vs-tallies.inp`

example was changed to see the new shape of the wavelength distributions (see attached).

And, with these implementations, now the peak of both distributions is at the same position